Two capacitors each of capacity C are connected as shown in the circuit across a battery. $W_{1}$ is the total energy stored in two capacitors when switch is closed. Now the switch S is opened and the two capacitors are immersed in a liquid of dielectric constant K, then the total energy stored in two capacitors is $W_{2}$ . The ratio of $W_{1} a n d W_{2}$ is

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$\frac{K}{K^{2} + 1}$

$\frac{2 K}{K^{2} + 1}$

$\frac{K}{K + 1}$

$\frac{K}{2 K + 1}$

answer is B.

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Detailed Solution

$\begin{array}{l} W_{1} = 2 U_{0} w h e r e U_{0} is energy in each capacitor when battery is connected \\ W_{2} = K U_{0} + \frac{U_{0}}{K} as one capacitor is disconnected from battery \\ \frac{W_{1}}{W_{2}} = \frac{2 U_{0}}{(K + \frac{1}{K}) U_{0}} = \frac{2 K}{K^{2} + 1} \end{array}$

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